Why is the domain of these two functions different? I tried graphing the functions $y = x^{.42}$ and $y = (x^{42})^{\frac{1}{100}}$. From my understanding of the laws of exponentiation these two expressions should be equivalent. What I found out after graphing these two functions was that they were equivalent for all non negative real numbers. The difference between these two functions was that the domain of $y = (x^{42})^{\frac{1}{100}}$ is $( -\infty, \infty)$ whereas the domain of $y = x^{.42}$ is $[0, \infty)$. I was wondering whether someone could explain why the domain of these two functions is different?
 A: Consider two functions: $f(x)=x^{\frac{1}{100}}$ and $g(x)=x^{42}$. Then $f$ has domain $[0, \infty)$ and $g$ has domain $\mathbb{R}$. Now consider the composition functions 
\begin{align*}
f(g(x))& =f(x^{42})=\left(x^{42}\right)^{\frac{1}{100}}\\
g(f(x))& =g(x^{\frac{1}{100}})=\left(x^{\frac{1}{100}}\right)^{42}
\end{align*}
The domain of $f(g(x))$ is $\mathbb{R}$ becuse $D_g=\mathbb{R}$ and its range is in $[0,\infty)$, whereas the domain of $g(f(x))$ has to be a subset of $D_f=[0,\infty)$. Hence the two compositions are different functions.
A: None of the existing answers have properly identified the problem:


*

*You must know exactly what definition of exponentiation you are using.


*There are at least two incompatible definitions that are commonly in use.


*Both of these two definitions satisfy $x^{ab} = (x^a)^b$ for positive real $x$ and rational $a,b$.


*No reasonable definitions can always satisfy $x^{ab} = (x^a)^b$ for negative $x$.

To show (4) it suffices to consider:

$-1 = (-1)^{1/5} = ((-1)^3)^{1/5} \overset{???}{=} (-1)^{3/5} = (-1)^{.6} = (-1)^{6/10} \overset{???}{=} ((-1)^6)^{1/10} = 1^{1/10} = 1$. (WRONG!!!)

Please refer to this post and this post for detailed explanation of two possible approaches to properly define exponentiation, as well as the two common but incompatible definitions for rational powers of negative reals.
A: You can calculate $x^{42}$ for any real number, and you'll get a non-negative number (because $42$ is even), and so you can take the $100^{\rm th}$ root. 
To calculate $x^{.42}$, it depends on your definition. You can notice that $.42=42/100$ and perform the previous calculation, and the domain will again be $(-\infty,\infty)$.
But in general, arbitrary powers are defined by 
$$
x^r:=e^{r\,\log x},
$$
which requires $x\geq0$. So, if you use this last definition, the domain would be $[0,\infty)$. 
A: For $b > 0$ then $b^{n/m}=\sqrt [m]{b^n} $ is well defined whereas the same definition for $b < 0$ is not.  If $b < 0$ for example, then $b^{3/2} = \sqrt {b^3} $ is not defined as $b^3$ is negative.  But $3/2 = 6/4$ makes $b^6$ positive so a 4th root is possible.  
So note: $.42=42/100 =21/50$ but $42/100$ is not in lowest terms.  This matters as $b^{21}$ will be negative if $b $ is but $b^{42} $ will be positive. 
So for non-negative $b $, $b^{n/m} = b^{2m/2n} $ but for negative $b $ that no longer holds.
